Shear and moment diagrams · Mechanics: Statics

Cut a beam anywhere and three internal forces hold the two halves together: the normal force $N$ along the beam, the shear force $V$ across it, and the bending moment $M$ , the couple that resists bending. Shear and moment diagrams plot $V$ and $M$ along the beam, and they are what you size a beam from.

Sign convention

A positive shear pushes the left part up and the right part down. A positive moment sags the beam (concave up, tension on the bottom); a negative moment hogs it (concave down). Pick one convention and hold it for the whole beam.

The relationships

Load, shear, and moment are tied together by calculus. Reading from the moment down:

V = \frac{dM}{dx}, \qquad q = \frac{dV}{dx} = \frac{d^2 M}{dx^2}

So the slope of the moment diagram is the shear, and the slope of the shear diagram is the load. Read it the other way and you integrate: the area under the load is the change in shear, and the area under the shear is the change in moment.

Two consequences worth memorising:

Where the shear crosses zero, the moment is at a maximum or minimum (its slope is zero there). That is how you locate the peak moment.
A point load makes the shear jump; a point moment makes the moment jump.

The order of lines

This is the shortcut that makes the diagrams fast. Each integration step raises the degree of the curve by one, so the shapes cascade:

Each integration step raises the degree by one. A triangular (linear) load starts one step higher, so its shear is parabolic and its moment is cubic.

A uniform load (a flat line) therefore gives a straight-sloped shear and a parabolic moment. Read the load and you already know the shape of the next two diagrams before computing a single number.

Worked example

A cantilever carrying a uniform load $w$ over its length $L$ . The load is constant, so the shear is linear and the moment is parabolic:

V(x) = w\,(L - x), \qquad M(x) = -\frac{w\,(L - x)^2}{2}

The shear is largest at the wall, $wL$ , and zero at the free tip; the moment is largest in magnitude at the wall, $-\tfrac{1}{2} w L^2$ , and zero at the tip. Notice that the shear is zero exactly where the moment flattens out.

Drawing them, step by step

Find the support reactions from equilibrium.
Build the shear from the left: drop by each point load, slope it by the distributed load (downward load slopes the shear down).
Build the moment from the left: its slope at every point is the current shear, so it climbs where $V > 0$ , falls where $V < 0$ , and peaks where $V = 0$ .
Use the order of lines to get each shape right, then pin the values with the areas.